# What is the perimeter of triangle with vertices of (1,2) (3,-4) and (-4,5)?

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Feb 28, 2016

Perimeter is $23.558$

#### Explanation:

To find the perimeter of a triangle with vertices of $\left(1 , 2\right)$, (3,−4) and (−4,5), we have to first find distance between each pair of points, which will give length of sides. For this we use distance formula between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$. Hence if lengths of sides are ${L}_{1} , {L}_{2} , {L}_{3}$, these are as follows:

${L}_{1} = \sqrt{{\left(3 - 1\right)}^{2} + {\left(\left(- 4\right) - \left(2\right)\right)}^{2}} = \sqrt{{2}^{2} + {\left(- 6\right)}^{2}} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} = 6.325$

${L}_{2} = \sqrt{{\left(- 4 - \left(3\right)\right)}^{2} + {\left(5 - \left(- 4\right)\right)}^{2}} = \sqrt{{\left(- 7\right)}^{2} + {9}^{2}} = \sqrt{49 + 81} = \sqrt{130} = 11.402$

${L}_{3} = \sqrt{{\left(- 4 - 1\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{{\left(- 5\right)}^{2} + {3}^{2}} = \sqrt{25 + 9} = \sqrt{34} = 5.831$

Hence Perimeter is $6.325 + 11.402 + 5.831 = 23.558$

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